# Torque during free fall

1. Oct 17, 2015

### UMath1

I am confused about whether objects when a nonuniform mass density experience net torque during free fall. If you take a rod that is heavier on one end and lighter on the other, should it not experience a net torque and angularly accelerate? But at the same time, the acceleration by gravity is uniform for the whole rod. And usually in the absence of air resistance it wouldn't rotate. It seems it only rotate if there is a pivot point. But I don't understand why.

2. Oct 17, 2015

### Staff: Mentor

Just because there's a net torque about a point does not necessarily mean that there will be an angular acceleration about that point. That "rule" applies for non-accelerating points or for the center of mass. I'm sure you'll agree that there's no net torque about the center of mass. And if you take torques about one end of the non-uniform rod, realize that that point is accelerating--it will turn out that the angular acceleration about that point will be zero.

3. Oct 17, 2015

### sophiecentaur

When the CM and the centre of pressure lie on a vertical line then there will be no torque. If the CM is vertically below the CP then the rod will be in stable equilibrium (there is a restoring torque for any displacement) and if the CM is above the CP, the equilibrium is unstable and the rod will soon start to rotate.

4. Oct 17, 2015

### Staff: Mentor

Why bring up center of pressure? The rod's in free fall.

5. Oct 17, 2015

### UMath1

I was considering the case where the center of mass is the axis of rotation. Then if the left side were denser, there would be a net counterclockwise torque? But it seems the axis of rotation must be at fulcrum for that to apply...but I can't explain to myself why that is.

6. Oct 17, 2015

### Staff: Mentor

No, there would be no net torque about the center of mass. If the left side of the rod were denser, the center of mass would be closer to that end.

Consider the definition of center of mass and how to calculate it. Compare that to calculating the net torque.

7. Oct 17, 2015

### sophiecentaur

I was throwing the rod out of an aeroplane (Woops!). That's the kind of 'free fall' that I thought of at first.

8. Oct 17, 2015

### UMath1

My bad...I mean the center of length of the rod.

9. Oct 17, 2015

### sophiecentaur

But if everything on the object is accelerating at g, there is no moment of torque of any point about any fulcrum you may choose.
accelerationrelative X mass = forcerelative
No relative acceleration produces no force.

10. Oct 17, 2015

### Staff: Mentor

You cannot take just any point as your axis, if that point is accelerating. (Unless it is the center of mass.) Is there a net torque about the center of the rod? Sure. Does that net torque lead to an angular acceleration? No.

11. Nov 7, 2015

### UMath1

so why is that you need a pivot or fixed point for angular acceleration to occur? Or do you? In the case of a ball rolling, its axis of rotation is not fixed.

12. Nov 7, 2015

### Staff: Mentor

You are making the assumption that torque about some axis always equals $I\alpha$. That's a reasonable assumption in intro physics, but it turns out not to be true except under special conditions. It's a topic covered in intermediate classical mechanics. (In intro physics just about all of the problems meet those special conditions, so you don't have to worry about it. Unless you're thinking ahead!)

One of those 'special conditions' is when the chosen point is the center of mass. In that case you can deduce that a net torque about such an axis does lead to an angular acceleration.

13. Nov 7, 2015

### UMath1

That actually helps answer another one of my questions. I was wondering why a cylinder wouldn't roll on a frictionless incline if gravity provided a torque if the axis were considered the point of contact.

Can you send me material that goes over this topic?

14. Nov 7, 2015

### Staff: Mentor

This is the sort of thing covered in just about any classical mechanics text. (If I find something online, I'll post a link.)

15. Nov 8, 2015

### elegysix

this is a bit off topic, but short answer - without friction there is no torque on the cylinder. it would slide down. the orientation doesn't matter.

16. Nov 8, 2015

### UMath1

No but if you consider the axis of rotation to be the point of contact with the ground, gravity does produce torque.

17. Nov 8, 2015

### Staff: Mentor

And that point of contact is accelerating, so things are a bit different when using that axis. The end result is the same, of course: No rotational acceleration about the center of mass.

18. Nov 8, 2015

### UMath1

The center of mass also accelerates

19. Nov 8, 2015

### Staff: Mentor

The center of mass has special properties. It doesn't matter if that point accelerates or not.

20. Nov 8, 2015

### UMath1

Right..thats what I want to know about. A text that goes over these special properties and the requirements for rotation.

21. Nov 8, 2015

### Staff: Mentor

22. Nov 8, 2015

### IgorIGP

UMath1, the CM coordinate is the result of an averaging. Ie for the gorisontal located linear body:
$x_c =\frac{1}{L} \int_0^L \rho _x (x)dx$
Suppose we have an inhomogeneous gorisontal rod. Obviously its center mass is not located in the middle. Let us mentally divide the rod through the center of mass by the the vertical plane. We will get the unequal geometrical parts whith an equal masses. Where will be the center mass each of them situated?
As I can see, your avatar tells that math is the thing you understand well.
So You can easily prove that the centers of mass of two geometricaly unequal parts obtained by the such dividing are situated at equal distances from the general CM (where the dividing plane located). So why have the torque be when we have two an equal masses located on equal distances from CM? That looks like balanced weighing-machine does not it?

Last edited: Nov 8, 2015